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Logarithm Test 1
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Logarithm Test 1

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  • Question 1/10
    1 / -0

    If log (x5 y2) = 5a + 2b and log (x2 y5) = 2a + 5b, find log (xy) in terms of a and b?

    Solutions

    log (x5 y2) = 5a + 2b

    ⇒ 5 logx + 2logy = 5a + 2b  → (1)

    Similarly, log (x2 y5) = 2a + 5b

    ⇒ 2 log x+ 5 log y = 2a + 5b → (2)

    ∴ log (xy) = log x + log y

    Adding (1) and (2) 7 log x + 7 log y = 7a + 7b

    log x + log y = a + b

  • Question 2/10
    1 / -0

    fraction numerator 1 over denominator log subscript 2 straight x end fraction plus fraction numerator 1 over denominator log subscript 2 straight x end fraction plus fraction numerator 1 over denominator log subscript 4 straight x end fraction plus fraction numerator 1 over denominator log subscript 5 straight x end fraction plus fraction numerator 1 over denominator log subscript 6 straight x end fraction plus fraction numerator 1 over denominator log subscript 7 straight x end fraction

    Solutions

    Similarly, 

    ∴ 

    = 

    = = logx 5040

  • Question 3/10
    1 / -0

    If logx 4 = 0.4, then the value of x is:

    Solutions

    logx 4 = 0.4 ⇔ logx 4 = ⇔x2/5 = 4 ⇔ x = 45/2 = (22)5/2

    ⇔ x = = 25 ⇔ x = 32.

  • Question 4/10
    1 / -0

     

    The value of log2 [log2 log2 log2 (65536)] is

    Solutions

    log2 log2 log2 log2 216 = log2 log2 log2 (16)

    = log2 log2 log2 (24) = log2 log2 (4)

    = log2 log2 (22) = log2 (2) = 1

  • Question 5/10
    1 / -0

    If the logarithm of 19683 to a base is 6, then what is the base?

    Solutions

    Let the base be x

    Logx 19683 = 6 ⇒ x6 = 19683 = (3)6

    ∴ x = 3

  • Question 6/10
    1 / -0

    If 5 + log10x = 5log10y, then express x in terms of y.

    Solutions

    5 + log10x = 5 log10y

    5 log1010 + log10x = log10y5 ⇒ log10x = log10y5 – log10105

    ⇒ log10x = log10  ⇒ x = y5/105 = (y/10)5

  • Question 7/10
    1 / -0

    The value of log5 is:

    Solutions

    Let log5  = n. Then, 5n = ⇒5n = 5–3 ⇒ n = – 3.

    ∴ log5  = –3.

  • Question 8/10
    1 / -0

    Find the logarithm of 1728 to the base 2 square root of 3 .

    Solutions

    let 1728 = x ⇒ (2)x = 1728

    ∵ 1728 = 26 ()6 = (2)6

     

    (2)x = (2)6

    On comparing ⇒ x = 6.

  • Question 9/10
    1 / -0

    Find the logarithm of 64 × cube root of 512 to the base cube root of 8

    Solutions

    = 

    = (3) (3) log88 = 9(1) = 9

  • Question 10/10
    1 / -0

    Simplify: log2 log2 log2 log subscript square root of 3 end subscript 6561.

    Solutions

    6151 =   = 16

    log2 16 = log224 = 4

    log24 = log222 = 2

    log22 = 1

    ∴ log2 log2 log2 log2 6561 = 1.

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